Question

For the values give for ΔH and ΔS, calculate ΔG foreach for each of the following reactions at 298 K. If the reactionis not spontaneous under standard conditions at 298 K, at whattemperature (if any) would the reaction become spontaneous?

a) 2PbS(s)+3O₂(g) ---> 2PbO(s)+2SO₂(g)
ΔH= -844kj ; ΔS = -165 J/K

b)2POCl₃(g)--->2PCl₃(g)+O₂(g)
ΔH= 572kJ ; ΔS= 179 J/K

Answer 1

Concepts and reason

Gibb’s free energy $\left( {\Delta G} \right)$ is a thermodynamic quantity which gives the maximum reversible work at constant temperature and pressure.

It is also explained mathematically as,

$\Delta G = \Delta H - T\Delta S$

Where, $\Delta H$ is the change in energy and $\Delta S$ is the change in entropy.

When the sign of $\Delta G$ is negative then the reaction is spontaneous and when the sign is positive then the reaction is non-spontaneous.

In the given question, you need to determine the value of $\Delta G$ for different reactions and then check the spontaneity of the reaction.

Fundamentals

The change in energy $\left( {\Delta H} \right)$ during phase change or during a reaction is known as Enthalpy. When energy is absorbed during reaction or phase change the sign for enthalpy is taken as positive and when energy is released during reaction or phase change the sign for enthalpy is taken as negative.

The change in entropy (degree of randomness, $\Delta S$ ) during phase change or during a reaction is known as Entropy. When the degree of randomness is increased, it is given a positive sign and when degree of randomness is decreased, it is given a negative sign.

a

The reaction is

${\rm{2PbS}}\left( s \right) + 3{{\rm{O}}_2}\left( g \right) \to 2{\rm{PbO}}\left( s \right) + 2{\rm{S}}{{\rm{O}}_{\rm{2}}}\left( g \right)$

The Gibb’s free energy at 298 K for the above reaction is as follows:

$\begin{array}{c}\\\Delta G = \Delta H - T\Delta S\\\\ = \left( { - 844{\rm{ kJ}}} \right) - \left( {298{\rm{ K}}} \right)\left( { - 165{\rm{ J/K}}} \right)\left( {\frac{{1{\rm{ kJ}}}}{{1000{\rm{ J}}}}} \right)\\\\ = - 794.83{\rm{ kJ}}\\\end{array}$

b

The reaction is

${\rm{2POC}}{{\rm{l}}_3}\left( g \right) \to 2{\rm{PC}}{{\rm{l}}_3}\left( g \right) + {{\rm{O}}_2}\left( g \right)$

The Gibb’s free energy at 298 K for the above reaction is as follows:

$\begin{array}{c}\\\Delta G = \Delta H - T\Delta S\\\\ = \left( {572{\rm{ kJ}}} \right) - \left( {298{\rm{ K}}} \right)\left( {179{\rm{ J/K}}} \right)\left( {\frac{{1{\rm{ kJ}}}}{{1000{\rm{ J}}}}} \right)\\\\ = 518.658{\rm{ kJ}}\\\end{array}$

The temperature at which Gibb’s free energy is negative for the above reaction is calculated as follows:

$\begin{array}{c}\\\Delta G < 0\\\\\Delta H - T\Delta S < 0\\\\\Delta H < T\Delta S\\\\\frac{{\Delta H}}{{\Delta S}} < T\\\\\frac{{572\;{\rm{kJ}}}}{{179\;{\rm{J/K}}\left( {\frac{{1\;{\rm{kJ}}}}{{1000\;{\rm{J}}}}} \right)}} < T\\\\3195.53\;{\rm{K}} < T\\\end{array}$

Ans: Part a

The Gibb’s free energy $\left( {\Delta G} \right)$ for the reaction is negative, therefore, the reaction is spontaneous.